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Chapter 3: Problem 49

A force, \(\mathbf{F}=(5 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mathbf{N}\) is applied over a particle which displaces it from its origin to the point \(\mathbf{r}=(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \mathbf{m}\). The work done on the particle in joule is (a) \(-\underline{7}\) (b) \(+7\) (c) \(+\underline{10}\) (d) \(+13\)

Short Answer

Expert verified
The work done is +7 joules.

Step by step solution

01

Understanding the Problem

First, we need to find the work done by force \( \mathbf{F} \) on a particle moving from the origin to the point \( \mathbf{r} = (2 \hat{\mathbf{i}} - \hat{\mathbf{j}}) \mathbf{m} \). The work done is calculated as the dot product of force and displacement vectors.
02

Identify Force and Displacement Vectors

We are given the force vector \( \mathbf{F} = 5 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} \), and the displacement vector is \( \mathbf{r} = 2 \hat{\mathbf{i}} - \hat{\mathbf{j}} \).
03

Calculate the Dot Product

The dot product is given by \( \mathbf{F} \cdot \mathbf{r} = (5 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) \cdot (2 \hat{\mathbf{i}} - \hat{\mathbf{j}}) \).
04

Compute the Components

Calculating the dot product: \[ 5 \times 2 + 3 \times (-1) + 2 \times 0 = 10 - 3 + 0 = 7 \]
05

Final Answer

The work done on the particle is \(+7\) joules. This matches option (b) \(+7\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Dot Product
The vector dot product is a crucial concept in physics and mathematics, particularly when calculating work done. The dot product, also known as the scalar product, combines two vectors to produce a scalar (a single number). It is denoted by a 'dot' between two vectors. To compute it, multiply the components of the vectors along the same direction and then sum these products.

In our exercise, the force vector (\( \mathbf{F} \)) and the displacement vector (\( \mathbf{r} \)) are used in the dot product calculation to find work.
  • Here, calculate using corresponding components: (\( 5 \times 2 \), (\( 3 \times -1 \)), and (\( 2 \times 0 \)).
  • The product yields the sum 10 - 3 + 0, which results in 7 joules.
This method is vital because it allows understanding of how vector directions contribute individually to work.
Force Vector
A force vector represents both the magnitude and direction of a force, which is essential in determining its effects. In the given exercise, the force vector is (\( \mathbf{F} = 5 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} \)). It means the force has components acting in three-dimensional space: x (\( \hat{\mathbf{i}} \)), y (\( \hat{\mathbf{j}} \)), and z (\( \hat{\mathbf{k}} \)) directions.

Each component's magnitude indicates how strongly the force acts in that particular direction.
  • Consider the force proteins of 5, 3, and 2 acting in x, y, and z, respectively.
  • This vector format uses unit vectors like (\( \hat{\mathbf{i}} \)) to provide a standard direction reference.
Understanding the force vector is critical not only for calculating work but also for predicting how objects will move when forces are applied.
Displacement Vector
The displacement vector shows the change in position of an object in space. It's essential to distinguish between distance and displacement. Displacement is a vector, meaning it has direction and magnitude, shown as (\( \mathbf{r} = 2 \hat{\mathbf{i}} - \hat{\mathbf{j}} \)) in our exercise.
  • Here, the object moves from the origin to the point (2, -1) in the x-y plane.
  • The components, 2 and -1, represent movement in the x and y directions respectively, showing positive and negative shifts.
This vector鈥檚 importance lies in its direct involvement in calculating the work through its dot product with force, revealing how movement in certain directions impacts physical processes.

Understanding displacement is crucial for anyone analyzing motion in physics.
Particle Motion
Particle motion applies the principles of vectors, and analyzing it requires understanding how particles (often simplified models of objects) react to applied forces. When a force vector affects a particle, it changes position, moving along a specified path given by a displacement vector.

The problem we solved demonstrates such motion.
  • The force vector shifts the particle from the origin to another point in the space.
  • This motion is both caused and described by the vectors we discussed: force and displacement.
This concept serves as a foundation for more complex physics topics, such as kinematics and dynamics, where motion under various conditions is analyzed. Understanding particle motion gives insight into the basic mechanics of how forces move objects.
Physics Problem Solving
Physics problem solving often requires breaking down complex scenarios into simpler parts, as shown in the exercise. To solve physics problems, it's essential to:
  • Understand the problem scenario, identifying knowns and unknowns.
  • Choose applicable physical laws or mathematical operations. For our solution, it was the dot product operation.
  • Execute computations step-by-step, ensuring each step logically follows the last.
Physics problem solving, as illustrated, is structured and methodical. This approach is applicable not only for textbook problems but also for real-world applications, helping to comprehend and predict phenomena. Building this habit can improve both academic and practical competency in physics.

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Most popular questions from this chapter

A body constrained to move in \(y\)-direction, is subjected to a force given by \(\mathbf{F}=(-2 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\) done by this force in moving the body through a distance of \(10 \mathrm{~m}\) along \(y\)-axis? (a) \(190 \mathrm{~J}\) (b) \(160 \mathrm{~J}\) (c) \(150 \mathrm{~J}\) (d) \(20 \mathrm{~J}\)

Consider a vector \(\mathbf{F}=4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}\). Another vector that is perpendicular to \(\mathbf{F}\) is (a) \(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) (b) \(6 \hat{\mathrm{j}}\) (c) \(7 \hat{j}\) (d) \(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}\)

The \(x\) and \(y\) components of a force are \(2 \mathrm{~N}\) and \(-3 \mathrm{~N}\). The force is (a) \(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}\) (b) \(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) (c) \(-2 \hat{i}-3 \hat{j}\) (d) \(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}\)

A proton in a cyclotron changes its velocity from \(30 \mathrm{kms}^{-1}\) north to \(40 \mathrm{kms}^{-1}\) east in \(20 \mathrm{~s}\). What is the average acceleration during this time (a) \(2.5 \mathrm{~km} \mathrm{~s}^{-2}\) at \(37^{\circ} \mathrm{E}\) of \(\mathrm{S}\) (b) \(2.5 \mathrm{~km} \mathrm{~s}^{-2}\) at \(37^{\circ} \mathrm{N}\) of \(\mathrm{E}\) (c) \(2.5 \mathrm{~km} \mathrm{~s}^{-2}\) at \(37^{\circ} \mathrm{N}\) of \(\mathrm{S}\) (d) \(2.5 \mathrm{~km} \mathrm{~s}^{-2}\) at \(37^{\circ} \mathrm{E}\) of \(\mathrm{N}\)

A particle is displaced from a position \((2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})\) to another position \((3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\) under the action of the force \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})\). The work done by the force is an arbitrary unit is (a) 8 (b) 10 (c) 12 (d) 16
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